Template:Example: Bayesian Test Design with Prior Information from Expert Opinion: Difference between revisions

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::<math> R=\text{BetaINV}\left(1-CL,\alpha\,\!,\beta\,\!\right)=0.902996 </math>  
::<math> R=\text{BetaINV}\left(1-CL,\alpha\,\!,\beta\,\!\right)=0.902996 </math>  


The confidence level ''CL'' can be found if system reliability ''R'', number of units ''n'', and number of failures ''r'' are known. Given the following data:


:*''R'' = 0.9
'''Solve for Confidence Level ''CL'''''
:*''n'' = 20
:*''r'' = 1


the number of successes ''s'' is:
Given ''R'' = 0.9, ''n'' = 20, and ''r'' = 1, using the above prior information on system reliability to solve for ''CL''.


::<math> s = n – r = 19 </math>
First, we get the number of successes ''s'' is:  


and the posterior distribution is calculated as:  
::<math> s = n – r = 19 </math>


::<math> \alpha\,\!=\alpha_{0}+s=21.76333 </math>
Then the parameters in the posterior Beta distribution for ''R'' are calculated as:  


::<math> \beta\,\!=\beta_{0}+r=1.44363 </math>  
::<math> \alpha\,\!=\alpha\,\!_{0}+s=21.76333 </math>
 
::<math> \beta\,\!=\beta\,\!_{0}+r=1.44363 </math>  
 
Finally, from this posterior distribution, the corresponding confidence level for reliablity ''R''=0.9 is:


::<math> CL=\text{BetaDist}\left(R,\alpha,\beta\right)=0.812164 </math>  
::<math> CL=\text{BetaDist}\left(R,\alpha,\beta\right)=0.812164 </math>  


The number of units ''n'' can be found if system reliability ''R'', confidence level ''CL'', and number of failures ''r'' are known. Given the following data:


:*''R'' = 0.9
'''Solve for Sample Size ''n'''''
:*''CL'' = 0.8
:*''r'' = 1
the '''Number of Units''' utility in the '''Non-Parametric Binomial''' tab of the '''Design a Reliability Demonstration Test''' window can be used to solve for ''n''.


[[Image:drt8.jpg|thumb|center|500px| ]]
Given ''R'' = 0.9, ''CL'' = 0.8, and ''r'' = 1, using the above prior information on system reliability to solve the required sample size in the demonstration test.


The figure above shows that, in this case, ''n'' = 28.925085. The posterior distribution can now be calculated as <math> s=n-r=27.925085 </math> <math> \alpha=\alpha_{0}+s=30.688416 </math> <math> \beta=\beta_{0}+r=1.44363 </math>
Again the above Beta distribution equation for the system reliability can be utilized. The figure in below shows the result from Weibull++.


which results in a confidence level of <math> CL=1-1-\text{BetaDist}\left(R,\alpha,\beta\right)=0.91775 </math> Since the confidence level (0.91775) is greater than that which is required (0.8), we can reduce the number of units ''n'' until the calculated confidence level is close to the required value of 0.8. This results in the number of units ''n'' = 19.31.
[[Image:drt8.jpg|thumb|center|500px| ]]

Revision as of 17:39, 23 February 2012

Bayesian Test Design with Prior Information from Expert Opinion

Suppose you wanted to know the reliability of a system and you had the following prior knowledge of the system:

  • Lowest possible reliability: a = 0.8
  • Most likely reliability: b = 0.85
  • Highest possible reliability: c = 0.97


This information can be used to approximate the expected value and the variance of the prior system reliability.

[math]\displaystyle{ E\left(R_{0}\right)=\frac{a+4b+c}{6}=0.861667 }[/math]
[math]\displaystyle{ Var\left(R_{0}\right)=\frac{c-a}{6}=0.028333 }[/math]


These approximations of the expected value and variance of the prior system reliability can then be used to estimate [math]\displaystyle{ \alpha_{0} }[/math] and [math]\displaystyle{ \beta_{0} }[/math] used in the Beta distribution for the system reliability, as given next:

[math]\displaystyle{ \alpha\,\!_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]=2.763331 }[/math]


[math]\displaystyle{ \beta\,\!_{0}=\left(1-E\left(R_{0}\right)\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]=0.44363 }[/math]

With [math]\displaystyle{ \alpha_{0} }[/math] and [math]\displaystyle{ \beta_{0} }[/math] known, any single value of the 4 quantities system reliability R, confidence level CL, number of units n, or number of failures r can be calculated from the other 3 using the Beta distribution function:

[math]\displaystyle{ 1-CL=\text{Beta}\left(R,\alpha,\beta\right)=\text{Beta}\left(R,n-r+\alpha_{0},r+\beta_{0}\right) }[/math]


Solve for System Reliability R

Given CL = 0.8, n = 20, and r = 1, using the above prior information to solve R.

First, we get the number of successes s is:

[math]\displaystyle{ s = n – r = 19 }[/math]

Then the parameters in the posterior Beta distribution for R are calculated as:

[math]\displaystyle{ \alpha\,\!=\alpha\,\!_{0}+s=21.76333 }[/math]
[math]\displaystyle{ \beta\,\!=\beta\,\!_{0}+r=1.44363 }[/math]

Finally, from this posterior distribution,the system reliability R at confidence level of CL=0.8 is solved as:

[math]\displaystyle{ R=\text{BetaINV}\left(1-CL,\alpha\,\!,\beta\,\!\right)=0.902996 }[/math]


Solve for Confidence Level CL

Given R = 0.9, n = 20, and r = 1, using the above prior information on system reliability to solve for CL.

First, we get the number of successes s is:

[math]\displaystyle{ s = n – r = 19 }[/math]

Then the parameters in the posterior Beta distribution for R are calculated as:

[math]\displaystyle{ \alpha\,\!=\alpha\,\!_{0}+s=21.76333 }[/math]
[math]\displaystyle{ \beta\,\!=\beta\,\!_{0}+r=1.44363 }[/math]

Finally, from this posterior distribution, the corresponding confidence level for reliablity R=0.9 is:

[math]\displaystyle{ CL=\text{BetaDist}\left(R,\alpha,\beta\right)=0.812164 }[/math]


Solve for Sample Size n

Given R = 0.9, CL = 0.8, and r = 1, using the above prior information on system reliability to solve the required sample size in the demonstration test.

Again the above Beta distribution equation for the system reliability can be utilized. The figure in below shows the result from Weibull++.

File:Drt8.jpg
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